The function f(x)=e^|x-1| is

The correct answer is Option (3) → continuous everywhere but not differentiable at \( x=1 \)

Since \( e^{u} \) and \( u=|x-1| \) are both continuous everywhere, their composition f is continuous everywhere. The only doubtful point for differentiability is x = 1, where |x − 1| has a corner.

    \( \displaystyle f(x)=\begin{cases}e^{1-x}, & x\lt 1\\e^{x-1}, & x\ge 1\end{cases} \)

    \( \displaystyle \text{LHD at }1=\lim_{x\to 1^-}\left(-e^{1-x}\right)=-1 \)

    \( \displaystyle \text{RHD at }1=\lim_{x\to 1^+}e^{x-1}=1 \)

The left-hand and right-hand derivatives differ, so f is not differentiable at x = 1. Elsewhere each piece is differentiable.

So f is continuous everywhere but not differentiable at x = 1.